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Ivanshal [37]
2 years ago
15

A rubber ball weighs 49 N.. a) What is the mass of the ball? (Answer: 5.0 kg). b) What is the acceleration of the ball if an upw

ard force of 69 N is applied? (Answer: 4.0 m/s/s). For a) I used the equation W = Fg = mg, which becomes 49 = m(9.8). This approximates m to 5.0 kg.
---
For b) I used the equation
ΣFy=Fa−W=ma
which becomes 69-49 = 20 N.
I then use equation F = ma which becomes 20 = 5.0a, which makes a
4.0m/s2
Physics
1 answer:
weqwewe [10]2 years ago
4 0
The answer you presented is correct.

For a) I used the equation W = Fg = mg, which becomes 49 = m(9.8). This approximates m to 5.0 kg.
<span>For b) I used the equation
</span><span>ΣFy=Fa−W=ma</span> which becomes 69-49 = 20 N.
I then use equation F = ma which becomes 20 = 5.0a, which makes a <span>4.0m/<span>s2</span></span>
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Answer:

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Explanation:

because it is

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3 years ago
A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.
melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

V_{f}^{2} = V_{0}^{2} + 2ad

<u>Where</u>:

V_{f}: is the final speed = 8.89 m/s

V_{0}: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m    

a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

a_{x} = a*cos(\theta)

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}

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I hope it helps you!

7 0
3 years ago
In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
lara31 [8.8K]

Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

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2 years ago
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<span>the body is moving horizontally, it doesnt matter watever kind of horizontal forces are acting.
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3 years ago
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If total charge Q is enclosed in the surface of cube

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so flux linked with each face is given by formula

\phi = \frac{Q}{6\epsilon_0}

so each face will have above flux due to central position of charge Q

3 0
2 years ago
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